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JOEL LEWIS: Hi.

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Welcome back to recitation.

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In lecture, you've been
learning about Stokes' Theorem,

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and I have a nice exercise on
Stokes' Theorem for you here.

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So I'm going to
let F be this field

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that I've written just above me.

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So it's 2x*z minus 2y comma 2y*z
plus 2x comma x square plus y

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square plus z square.

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And I've got C.

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So C is this
complicated-looking curve here.

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So it sort of dips up
and down and back around.

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But the thing I'm
really going to tell you

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about it is that it all lies
on this cylinder of radius b.

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So C is this curve in
the cylinder of radius b

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that wraps around it once,
but behaves kind of oddly

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while it's wrapping around.

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So what I'd like
you to do is I'd

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like you to use Stokes'
Theorem to compute

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the integral around this
curve of F dot dr. Now,

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my hint to you is that
for Stokes' Theorem,

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you can use-- just like you
have for Green's Theorem

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and for Divergence Theorem
that we've talked about before,

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you have these
extended versions that

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let you consider more
than one boundary piece.

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So the same thing works
for Stokes' Theorem.

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So Stokes' Theorem
works perfectly well

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when you have a piece
of a surface with more

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than one boundary curve,
provided you orient everything

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correctly.

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So you might think about how
you can use Stokes' Theorem

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to replace this complicated
curve with a surface integral

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and an easier to
understand curve.

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And if you can do that,
then computing the other two

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gives you the third one.

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All right.

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So that's my hint to you
for computing this integral.

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So why don't you pause the
video, have a go at that,

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come back, and we can
work on it together.

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Hopefully, you had some luck
working on this problem.

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Let's talk about it.

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So before I left, I
gave you this hint

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that maybe the thing to do here
isn't to try and parametrize

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this curve directly and compute
the line integral directly

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since it's a
complicated-looking curve,

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and also since I haven't really
given you enough information

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to do that, and instead to think
about applying Stokes' Theorem.

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So to think about applying
Stokes' Theorem, what we'd like

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is a nice surface, with this
curve as part of its boundary.

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Well, what is such a surface?

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Well, this curve lies all
on the cylinder of radius b.

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So a natural choice
for a surface

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is to use some piece
of this cylinder.

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So maybe we could use the
piece of this cylinder

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with this as its upper boundary.

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So then what might be a natural
lower boundary to choose?

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Well, we just want to choose
something nice and simple.

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Right?

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So, what's nice and simple?

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Well, maybe we can
choose this bottom circle

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that's in the plane y equals x.

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All right.

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So I'm going to call
that circle C_1.

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So that's the circle of
radius b in the xy-plane.

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Sorry, not the plane y equals x.

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The xy-plane.

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The plane z equals 0.

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So we've got the top curve C
and we've got this bottom curve

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C_1.

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Now, the way I've
oriented them, I've

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oriented them both so that
they're going counterclockwise

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as you look down
from the z-axis.

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So in that case, what
does Stokes' Theorem say?

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Well, Stokes' Theorem
says that the integral

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over the piece of the
surface between them--

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let's call it S-- of curl F
dot n with respect to surface

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area is equal to-- OK.

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So let's say we can give it the
outward pointing normal, say.

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In which case, C_1 will
be positively oriented

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and C will be
negatively oriented.

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So this is equal to
the line integral

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over C_1 of F dot dr minus
the line integral over C

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of F dot dr.

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And so what's nice
about this formula

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is that it replaces computing
the integral that we want.

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Instead of computing that,
we can try and compute

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this other line integral
and this surface integral.

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And if these are
easier to compute,

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then computing the two of them
gives us what the value of this

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is just by subtracting, or
by adding and subtracting,

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or whatever.

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By arithmetic, right?

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So if these integrals
are easy to compute,

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then that makes this one
easy without actually having

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to parametrize and compute it.

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So let's take a look at
what these integrals are.

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Let's do the surface integral
first since it's on the left.

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So in order to compute
the surface integral,

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we're going to need to
compute the curl of F.

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So OK.

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So F is this kind of
messy-looking thing here.

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So curl of F, well,
what have we got?

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So it's going to be
big thing times i hat.

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So it's going to be i hat
times this determinant, right?

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So let me write the determinant.

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So on top we've got
i hat, j hat, k hat,

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then we have the partial
x, partial y, partial z,

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and then we have the components.

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So these are 2x*z minus
2y, and 2y*z plus 2x,

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and x square plus y
square plus z square.

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All right.

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So that's i hat,
j hat, and k hat.

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And then we've got
partial over partial x,

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partial over partial y,
and partial over partial z.

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So this is what the
curl is, and so now we

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have to expand this out.

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So for i, it's going to be
partial y of x squared plus y

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squared plus z
squared-- so that's 2y--

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minus partial z of 2y*z plus
2x, so that's minus 2y, i hat.

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Plus-- for j hat, it's going to
be partial z of 2x*z minus 2y,

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so that's 2x-- minus partial
x of x squared plus y squared

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plus z squared, so
that's minus 2x, j hat.

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Plus-- for k hat, we want
partial x of 2y*z plus 2x,

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so that's 2-- minus
partial y of 2x*z minus 2y,

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so that's minus minus 2,
so that's plus 2, k hat.

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Oh.

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All right.

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OK.

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So the i-component is 0
and the j-component is 0.

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So this is a nice, simple one.

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So the curl here, the
k-component is just 4.

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So this is equal to 4k.

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OK.

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So that's what the curl of F is.

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Now what do we need to compute?

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We need to compute curl of F dot
the normal vector with respect

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to surface area.

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Now let's look at
what our surface is.

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Our surface is right here, and
it's this vertical cylinder.

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Well, what is the normal
vector of a vertical cylinder?

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Well, it's pointing
straight away from the axis.

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Right?

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It's perpendicular to the
surface of the cylinder,

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so it's parallel
to the xy-plane.

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It rotates as you go
around the cylinder,

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but it's always in the xy-plane.

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So what does that mean?

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Well, that means
in particular, it's

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perpendicular to things
in the z-direction.

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Right?

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So if we look, we see our curl
here is just straight upward

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in the z-direction.

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And our normal vector
has no z-component.

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It's only in the xy-plane.

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So this k hat is
orthogonal to n, OK?

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So the curl and
n are orthogonal.

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So their dot product is 0.

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So this surface integral
is a surface integral of 0.

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So it just gives you 0.

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OK.

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So great.

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So that's really nice.

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That simplifies
our life very much.

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Now, our line
integral that we want.

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We just have it in terms of
this one other line integral.

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Right?

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So the surface integral is 0.

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And let me see.

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Where should I put this?

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OK, the curl is 4k, so the
surface integral curl F dot n

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dS is also equal to 0.

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So having made that
simplification,

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now we just need
this other integral.

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We need this line
integral over C_1.

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And that'll give
us what we need.

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So let's have a go at that.

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So C_1 is the circle
of radius b centered

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at the origin in the xy-plane.

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OK.

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I'm not going to
write that down.

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I'm just going to say it.

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The circle of radius b centered
at the origin in the xy-plane.

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So OK.

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So it's not that
hard to parametrize.

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So it's parametrized by
x equals b cosine theta,

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y equals b sine theta.

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We should check.

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We should double-check
that we're

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doing the right direction
of parametrization.

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Let's go have a look.

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Let's see.

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Yes.

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OK.

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So we parametrized
this circle going

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counterclockwise
in the xy-plane.

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So good.

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So this parametrization is
going the right direction.

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Otherwise, we'd have to change
the sign of theta or something.

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So it's x is b cosine
theta, y is b sine theta.

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And we're going once
around the circle,

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so we want 0 less than or equal
to theta less than or equal

201
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to 2*pi.

202
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And so what do we have?

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So now, we want to
compute the integral

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over the circle of F dot
dr. So let's see what

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F looks like in this situation.

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So let's go back and look at
the expression for F over here.

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So in this plane, we
have z is equal to 0.

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So F is minus 2y, plus 2x,
x squared plus y squared.

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OK?

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OK, so let's come back then.

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00:11:03,600 --> 00:11:07,590
So F is what I
just said, so this

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is equal to the integral
over C of minus 2y dx,

213
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plus 2x dy-- plus x
squared plus y squared dz,

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but we're in the
plane z equals 0,

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so dz is always
0 in that plane--

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so we don't have a
third term there.

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Great.

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So this is our
integral, and now we

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can substitute from our
parametrization here.

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So this is equal to the
integral from 0 to 2*pi.

221
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So minus 2y dx.

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So that's minus 2b
sine theta, times-- dx

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is minus b sine theta
d theta-- plus 2x-- so

224
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that's 2b cosine theta-- times
dy, which is b cosine theta d

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theta.

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Whew.

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00:12:23,850 --> 00:12:26,665
This is quite a long
equation, isn't it?

228
00:12:26,665 --> 00:12:29,050
Or a long expression, I guess.

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So our line integral around
C of F dot dr is equal

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to the integral from 0 to 2*pi
of minus 2b sine theta times

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00:12:38,560 --> 00:12:39,960
minus b sine theta d theta.

232
00:12:39,960 --> 00:12:44,230
So this is 2b squared sine
squared theta d theta.

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And this is 2b cosine
squared theta d theta.

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So OK.

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So that 2 b squared
is a constant.

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00:12:50,660 --> 00:12:52,010
We can just factor it out.

237
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And we're left with sine squared
theta plus cosine squared theta

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d theta.

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All right.

240
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OK.

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That's great.

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I'm happy to have that.

243
00:13:00,500 --> 00:13:01,000
Right?

244
00:13:01,000 --> 00:13:02,583
Sine squared theta
plus cosine squared

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00:13:02,583 --> 00:13:03,950
theta, that's going to be 1.

246
00:13:03,950 --> 00:13:04,610
OK.

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So we can rewrite this.

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I'm going to bring
it back up here.

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So that's equal to the integral
from 0 to 2*pi of 2 b squared d

250
00:13:16,250 --> 00:13:22,650
theta, which is 4*pi b squared.

251
00:13:22,650 --> 00:13:23,330
Great.

252
00:13:23,330 --> 00:13:29,401
OK, so that's our line integral
around this bottom curve C.

253
00:13:29,401 --> 00:13:29,900
Oh, dear.

254
00:13:29,900 --> 00:13:33,630
I've been writing C, but this
is not our original curve C,

255
00:13:33,630 --> 00:13:36,761
this is our new curve
C_1, like I wrote there.

256
00:13:36,761 --> 00:13:37,260
Sorry.

257
00:13:37,260 --> 00:13:39,557
So everywhere I wrote the
line integral over C--

258
00:13:39,557 --> 00:13:41,140
both of these places--
it was supposed

259
00:13:41,140 --> 00:13:43,300
to be a line integral over C_1.

260
00:13:43,300 --> 00:13:45,260
Sorry about that.

261
00:13:45,260 --> 00:13:48,020
So we've got this line
integral over C_1,

262
00:13:48,020 --> 00:13:51,590
and it worked out
to 4*pi b squared,

263
00:13:51,590 --> 00:13:56,040
just using our usual
parametrize-and-compute

264
00:13:56,040 --> 00:13:58,110
technique for computing
line integrals.

265
00:13:58,110 --> 00:13:58,610
So OK.

266
00:13:58,610 --> 00:14:00,020
So now, let's see
where we're at.

267
00:14:00,020 --> 00:14:03,640
Let's go back over here
to when we wrote down

268
00:14:03,640 --> 00:14:06,270
what the extended Stokes'
Theorem says in our case.

269
00:14:06,270 --> 00:14:08,962
So Stokes' Theorem
told us that the thing

270
00:14:08,962 --> 00:14:10,670
we were interested
in-- this is the thing

271
00:14:10,670 --> 00:14:12,160
we're trying to compute, right?

272
00:14:12,160 --> 00:14:15,340
The problem asked us to
compute the line integral

273
00:14:15,340 --> 00:14:19,460
over C of F dot
dr. Well, extended

274
00:14:19,460 --> 00:14:21,830
Stokes' Theorem said, in
order to compute this line

275
00:14:21,830 --> 00:14:24,564
integral, what you can do is
you can compute this surface

276
00:14:24,564 --> 00:14:26,480
integral over S, and you
can compute this line

277
00:14:26,480 --> 00:14:28,780
integral over this
other curve C_1,

278
00:14:28,780 --> 00:14:31,970
and then these three things have
to satisfy this relationship.

279
00:14:31,970 --> 00:14:33,780
That's what's
Stokes' Theorem says.

280
00:14:33,780 --> 00:14:34,800
And now we've computed.

281
00:14:34,800 --> 00:14:36,300
We've computed the
surface integral,

282
00:14:36,300 --> 00:14:40,684
and we found it was equal to 0
by a simple geometric argument

283
00:14:40,684 --> 00:14:42,850
that didn't require us to
actually compute a surface

284
00:14:42,850 --> 00:14:44,560
integral.

285
00:14:44,560 --> 00:14:48,110
And we computed this
line integral, just now,

286
00:14:48,110 --> 00:14:49,760
by parametrizing
and computing it.

287
00:14:49,760 --> 00:14:50,260
So OK.

288
00:14:50,260 --> 00:14:55,020
So this was 0 and this
was 4*pi b squared.

289
00:14:55,020 --> 00:14:58,570
So if we just add our
integral in question

290
00:14:58,570 --> 00:15:00,510
to the other side,
what we find--

291
00:15:00,510 --> 00:15:04,360
I'm going to go find some empty
board space to write it down--

292
00:15:04,360 --> 00:15:14,780
so our integral, the
integral over C of F

293
00:15:14,780 --> 00:15:22,785
dot dr is equal to this other
line integral minus the surface

294
00:15:22,785 --> 00:15:23,285
integral.

295
00:15:23,285 --> 00:15:27,864
So it's equal to 4*pi
b squared minus 0.

296
00:15:27,864 --> 00:15:30,280
Just rearranging that equation
we were looking at a second

297
00:15:30,280 --> 00:15:31,550
ago from Stokes' Theorem.

298
00:15:31,550 --> 00:15:36,390
So it's just 4*pi b squared.

299
00:15:36,390 --> 00:15:39,160
So that's the answer,
and I'll end there.